C Examples

C Program - Find all factors of a Number



Objective: Write a C program to find all distinct factors (divisors) of a given natural number. The divisors of few numbers are given below:

 Number: 10
 Divisors: 1 2 5 10

 Number: 15
 Divisors: 1 3 5 15

 Number: 100
 Divisors: 1 2 4 5 10 20 25 50 100 

Method 1: Using iteration

One of the basic approach is to iterate from 1 to n and in each iteration check whether the number divides n. If it divides then print it.

#include <stdio.h>

//function to print all divisors of a number
static void printDivisors(int n) {
  printf("Divisors of %i are: ", n);
  for(int i = 1; i <= n; i++) {
    if(n%i == 0)
      printf("%i ", i);
  }
  printf("\n");
}

int main(){
  printDivisors(10);
  printDivisors(50);
  printDivisors(100);
  return 0;
}

The above code will give the following output:

Divisors of 10 are: 1 2 5 10 
Divisors of 50 are: 1 2 5 10 25 50 
Divisors of 100 are: 1 2 4 5 10 20 25 50 100 

Method 2: Optimized Code

Instead of checking the divisibility of the given number from 1 to n, it is checked till square root of n. For a factor larger than square root of n, there must the a smaller factor which is already checked in the range of 1 to square root of n.

#include <stdio.h>
#include <math.h>

//function to print all divisors of a number
static void printDivisors(int n) {
  printf("Divisors of %i are: ", n);
  //loop from 1 to sqrt(n)
  for(int i = 1; i <= sqrt(n); i++) {
    if(n%i == 0) {
      if(n/i == i)
        printf("%i ", i);
      else
        printf("%i %i ", i, n/i);
    }
  }
  printf("\n");
}

int main(){
  printDivisors(10);
  printDivisors(50);
  printDivisors(100);
  return 0;
}

The above code will give the following output:

Divisors of 10 are: 1 10 2 5 
Divisors of 50 are: 1 50 2 25 5 10 
Divisors of 100 are: 1 100 2 50 4 25 5 20 10 

Method 3: Optimized Code with sorted result

In the previous method, results are produced in a irregular fashion (printed in pairs - small number and large number). The result can be sorted by storing the larger number and print them later on. Consider the example below:

#include <stdio.h>
#include <math.h>

//function to print all divisors of a number
static void printDivisors(int n) {
  printf("Divisors of %i are: ", n);
  
  //creating an array to store larger numbers
  int arr[n];
  int j = 0;
  
  //loop from 1 to sqrt(n)
  for(int i = 1; i <= sqrt(n); i++) {
    if(n%i == 0) {
      if(n/i == i)
        printf("%i ", i);
      else {
        printf("%i ", i);
        //storing the large number of a pair
        arr[j++] = (n/i);        
      }
    }
  }

  //printing stored large numbers of pairs
  for(int i = j - 1; i >= 0; i--)
    printf("%i ", arr[i]);
  printf("\n");
}

int main(){
  printDivisors(10);
  printDivisors(50);
  printDivisors(100);
  return 0;
}

The above code will give the following output:

Divisors of 10 are: 1 2 5 10 
Divisors of 50 are: 1 2 5 10 25 50 
Divisors of 100 are: 1 2 4 5 10 20 25 50 100 

Method 4: Another Optimized Code

To produce the result in sorted order, we can iterate from 1 to square root of n and printing the number which divides n. After that we can iterate back (in reverse order) and printing the quotient of all numbers which divides n.

#include <stdio.h>
#include <math.h>

//function to print all divisors of a number
static void printDivisors(int n) {
  printf("Divisors of %i are: ", n);
  //loop from 1 to sqrt(n)
  int i;
  for(i = 1; i <= sqrt(n); i++) {
    if(n%i == 0) 
      printf("%i ", i);
    
    //handing perfect squares
    if(n/i == i) {
      i--; break;
    }
  }

  for(; i >= 1; i--) {
    if(n%i == 0) 
      printf("%i ", n/i);
  }  
  printf("\n");
}

int main(){
  printDivisors(10);
  printDivisors(50);
  printDivisors(100);
  return 0;
}

The above code will give the following output:

Divisors of 10 are: 1 2 5 10 
Divisors of 50 are: 1 2 5 10 25 50 
Divisors of 100 are: 1 2 4 5 10 20 25 50 100 




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